Theorem sban 1232

Description: Conjunction inside and outside of a substitution are equivalent.

Assertion

Ref	Expression
sban	|- ([y / x](ph /\ ps) <-> ([y / x]ph /\ [y / x]ps))

Proof of Theorem sban

Step	Hyp	Ref	Expression
1		sbn 1226	. . 3 |- ([y / x] -. (ph -> -. ps) <-> -. [y / x](ph -> -. ps))
2		sbim 1229	. . . . 5 |- ([y / x](ph -> -. ps) <-> ([y / x]ph -> [y / x] -. ps))
3		sbn 1226	. . . . . 6 |- ([y / x] -. ps <-> -. [y / x]ps)
4	3	imbi2i 185	. . . . 5 |- (([y / x]ph -> [y / x] -. ps) <-> ([y / x]ph -> -. [y / x]ps))
5	2, 4	bitr 173	. . . 4 |- ([y / x](ph -> -. ps) <-> ([y / x]ph -> -. [y / x]ps))
6	5	negbii 187	. . 3 |- (-. [y / x](ph -> -. ps) <-> -. ([y / x]ph -> -. [y / x]ps))
7	1, 6	bitr 173	. 2 |- ([y / x] -. (ph -> -. ps) <-> -. ([y / x]ph -> -. [y / x]ps))
8		df-an 225	. . 3 |- ((ph /\ ps) <-> -. (ph -> -. ps))
9	8	sbbii 1170	. 2 |- ([y / x](ph /\ ps) <-> [y / x] -. (ph -> -. ps))
10		df-an 225	. 2 |- (([y / x]ph /\ [y / x]ps) <-> -. ([y / x]ph -> -. [y / x]ps))
11	7, 9, 10	3bitr4 183	1 |- ([y / x](ph /\ ps) <-> ([y / x]ph /\ [y / x]ps))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146   /\ wa 223  [wsbc 1166

This theorem is referenced by:  sb3an 1233  sbbi 1234  sbabel 1576  sbcang 1961  inab 2258  exss 2759  inopab 3258

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 959  ax-gen 960  ax-10 963  ax-12 965  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-10o 1136  ax-11o 1213

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 978  df-sb 1168

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